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# hoffman and kunze solution

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```Linear Algebra
Hoffman & Kunze

2nd edition

Answers and Solutions to Problems and Exercises
Typos, comments and etc...

Gregory R. Grant
University of Pennsylvania
email: ggrant@upenn.edu

Julyl 2017

2
Note

This is one of the ultimate classic textbooks in mathematics. It will probably be read for generations to come. Yet I
cannot find a comprehensive set of solutions online. Nor can I find lists of typos. Since I’m going through this book in some
detail I figured I’d commit some of my thoughts and solutions to the public domain so others may have an easier time than I
have finding supporting matericals for this book. Any book this classic should have such supporting materials readily avaiable.

If you find any mistakes in these notes, please do let me know at one of these email addresses:

ggrant543@gmail.com or greg@grant.org or ggrant@upenn.edu

The latex source file for this document is available at http://greg.grant.org/hoffman and kunze.tex. Use it wisely.
And if you cannot manage to download the link because you included the period at the end of the sentence, then maybe you
are not smart enough to be reading this book in the first place.

Chapter 1: Linear Equations

Section 1.1: Fields
Page 3. Hoffman and Kunze comment that the term “characteristic zero” is “strange.” But the characteristic is the smallest n
such that n · 1 = 0. In a characteristic zero field the smallest such n is 0. This must be why they use the term “characteristic
zero” and it doesn’t seem that strange.

Section 1.2: Systems of Linear Equations
Page 5 Clarification: In Exercise 6 of this section they ask us to show, in the special case of two equations and two unknowns,
that two homogeneous linear systems have the exact same solutions then they have the same row-reduced echelon form (we
know the converse is always true by Theorem 3, page 7). Later in Exercise 10 of section 1.4 they ask us to prove it when
there are two equations and three unknowns. But they never tell us whether this is true in general (for abitrary numbers of
unknowns and equations). In fact is is true in general. This explanation was given on math.stackexchange:

Solutions to the homogeneous system associated with a matrix is the same as determining the null space of the
relevant matrix. The row space of a matrix is complementary to the null space. This is true not only for inner
product spaces, and can be proved using the theory of non-degenerate symmetric bilinear forms.

So if two matrices of the same order have exactly the same null space, they must also have exactly the same row
space. In the row reduced echelon form the nonzero rows form a basis for the row space of the original matrix,
and hence two matrices with the same row space will have the same row reduced echelon form.

Exercise 1: Verify that the set of complex numbers described in Example 4 is a subfield of C.

Solution: Let F = {x + y√2 | x, y ∈ Q}. Then we must show six things:

1. 0 is in F

2. 1 is in F

3. If x and y are in F then so is x + y

4. If x is in F then so is −x
5. If x and y are in F then so is xy

6. If x , 0 is in F then so is x−1

For 1, take x = y = 0. For 2, take x = 1, y = 0. For 3, suppose x = a+b
√

2 and y = c+d
√

2. Then x+y = (a+c)+(b+d)
√

2 ∈ F.
For 4, suppose x = a + b

√
2. Then −x = (−a) + (−b)√2 ∈ F. For 5, suppose x = a + b√2 and y = c + d√2. Then

1

2 Chapter 1: Linear Equations

xy = (a + b
√

2)(c + d
√

2) = (ac + 2bd) + (ad + bc)
√

2 ∈ F. For 6, suppose x = a + b√2 where at least one of a or b is not
zero. Let n = a2 − 2b2. Let y = a/n + (−b/n)√2 ∈ F. Then xy = 1n (a + b

√
2)(a − b√2) = 1n (a2 − 2b2) = 1. Thus y = x−1 and

y ∈ F.

Exercise 2: Let F be the field of complex numbers. Are the following two systems of linear equations equivalent? If so,
express each equation in each system as a linear combination of the equations in the other system.

x1 − x2 = 0 3x1 + x2 = 0
2x1 + x2 = 0 x1 + x2 = 0

Solution: Yes the two systems are equivalent. We show this by writing each equation of the first system in terms of the
second, and conversely.

3x1 + x2 =
1
3

(x1 − x2) + 43(2x1 + x2)

x1 + x2 =
−1
3

(x1 − x2) + 23(2x1 + x2)
x1 − x2 = (3x1 + x2) − 2(x1 + x2)

2x1 + x2 =
1
2

(3x1 + x2) +
1
2

(x1 + x2)

Exercise 3: Test the following systems of equations as in Exercise 2.

−x1 + x2 +4x3 = 0 x1 − x3= 0
x1 + 3x2+8x3 = 0 x2 + x3= 0

1
2 x1 + x2 + 52 x3 = 0

Solution: Yes the two systems are equivalent. We show this by writing each equation of the first system in terms of the
second, and conversely.

x1 − x3 = −34 (−x1 + x2 + 4x3) + 14 (x1 + 3x3 + 8x3)
x2 + 3x3 = 14 (−x1 + x2 + 4x3) + 14 (x1 + 3x3 + 8x3)

and

−x1 + x2 + 4x3 = −(x1 − x3) + (x2 + 3x3)
x1 + 3x2 + 8x3 = (x1 − x3) + 3(x2 + 3x3)
1
2 x1 + x2 + 52 x3 = 12 (x1 − x3) + (x2 + 3x3)

Exercise 4: Test the following systems as in Exercie 2.

2x1 + (−1 + i)x2 + x4 = 0 (1 + i2 )x1+8x2 − ix3 −x4 = 0
3x2 − 2ix3 + 5x4 = 0 23 x1 − 12 x2 + x3 +7x4 = 0

Solution: These systems are not equivalent. Call the two equations in the first system E1 and E2 and the equations in the
second system E′1 and E

′
2. Then if E

′
2 = aE1 + bE2 since E2 does not have x1 we must have a = 1/3. But then to get the

coefficient of x4 we’d need 7x4 = 13 x4 + 5bx4. That forces b =
4
3 . But if a =

1
3 and b =

4
3 then the coefficient of x3 would have

to be −2i 43 which does not equal 1. Therefore the systems cannot be equivalent.

Exercise 5: Let F be a set which contains exactly two elements, 0 and 1. Define an addition and multiplication by the tables:

+ 0 1
0 0 1
1 1 0

· 0 1
0 0 0
0 0 1

Section 1.2: Systems of Linear Equations 3

Solution: We must check the nine conditions on pages 1-2:

1. An operation is commutative if the table is symmetric across the diagonal that goes from the top left to the bottom right.
This is true for the addition table so addition is commutative.
2. There are eight cases. But if x = y = z = 0 or x = y = z = 1 then it is obvious. So there are six non-trivial cases. If there’s
exactly one 1 and two 0’s then both sides equal 1. If there are exactly two 1’s and one 0 then both sides equal 0. So addition
is associative.
3. By inspection of the addition table, the element called 0 indeed acts like a zero, it has no effect when added to another
element.
4. 1 + 1 = 0 so the additive inverse of 1 is 1. And 0 + 0 = 0 so the additive inverse of 0 is 0. In other words −1 = 1 and
−0 = 0. So every element has an additive inverse.
5. As stated in 1, an operation is commutative if the table is symmetric across the diagonal that goes from the top left to the
bottom right. This is true for the multiplication table so multiplication is commutative.
6. As with addition, there are eight cases. If x = y = z = 1 then it is obvious. Otherwise at least one of x, y or z must equal 0.
In this case both x(yz) and (xy)z equal zero. Thus multiplication is associative.
7. By inspection of the multiplication table, the element called 1 indeed acts like a one, it has no effect when multiplied to
another element.
8. There is only one non-zero element, 1. And 1 · 1 = 1. So 1 has a multiplicative inverse. In other words 1−1 = 1.
9. There are eight cases. If x = 0 then clearly both sides equal zero. That takes care of four cases. If all three x = y = z = 1
then it is obvious. So we are down to three cases. If x = 1 and y = z = 0 then both sides are zero. So we’re down to the two
cases where x = 1 and one of y or z equals 1 and the other equals 0. In this case both sides equal 1. So x(y + z) = (x + y)z in
all eight cases.

Exercise 6: Prove that if two homogeneous systems of linear equations in two unknowns have the same solutions, then they
are equivalent.

Solution: Write the two systems as follows:

a11x + a12y = 0
a21x + a22y = 0

...
am1x + am2y = 0

b11x + b12y = 0
b21x + b22y```